<!DOCTYPE html><html lang="zh-CN" data-theme="light"><head><meta charset="UTF-8"><meta http-equiv="X-UA-Compatible" content="IE=edge"><meta name="viewport" content="width=device-width,initial-scale=1"><title>超几何分布抽取概率证明 | 云玩家</title><meta name="keywords" content="数学,分布,概率论"><meta name="author" content="云玩家"><meta name="copyright" content="云玩家"><meta name="format-detection" content="telephone=no"><meta name="theme-color" content="#ffffff"><meta name="description" content="简介相信大家都在高中接触过超几何分布. 超几何分布是不放回的, 该分布给出了抽取 $n$ 次后抽到次品次数的概率. 如果一个随机变量 $X$ 服从于超几何分布, 那么有$$P(X&#x3D;k)&#x3D;\frac{C_{N}^kC_{M-N}^{n-k}}{C_{M}^{n}}$$其中 $k$ 为抽取到的次品数, $M$ 为总数, $N$ 为总次品数. 问题生活中很多变量都服从于超几何分布, 我们举个例子  假设">
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fa-fw"></i></a></div></div></nav><div id="post-info"><h1 class="post-title">超几何分布抽取概率证明</h1><div id="post-meta"><div class="meta-firstline"><span class="post-meta-date"><i class="far fa-calendar-alt fa-fw post-meta-icon"></i><span class="post-meta-label">发表于</span><time class="post-meta-date-created" datetime="2020-05-23T09:12:52.000Z" title="发表于 2020-05-23 17:12:52">2020-05-23</time><span class="post-meta-separator">|</span><i class="fas fa-history fa-fw post-meta-icon"></i><span class="post-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-08-14T04:39:34.426Z" title="更新于 2021-08-14 12:39:34">2021-08-14</time></span><span class="post-meta-categories"><span class="post-meta-separator">|</span><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/%E6%95%B0%E5%AD%A6/">数学</a><i class="fas fa-angle-right post-meta-separator"></i><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/%E6%95%B0%E5%AD%A6/%E6%A6%82%E7%8E%87%E8%AE%BA/">概率论</a></span></div><div class="meta-secondline"><span class="post-meta-separator">|</span><span class="post-meta-pv-cv"><i class="far fa-eye fa-fw post-meta-icon"></i><span class="post-meta-label">阅读量:</span><span id="busuanzi_value_page_pv"></span></span><span class="post-meta-separator">|</span><span class="post-meta-commentcount"><i class="far fa-comments fa-fw post-meta-icon"></i><span class="post-meta-label">评论数:</span><a href="/math/probability_theory/hypergeometric_distribution/#post-comment"><span id="twikoo-count"></span></a></span></div></div></div></header><main class="layout" id="content-inner"><div id="post"><article class="post-content" id="article-container"><h1 id="简介"><a href="#简介" class="headerlink" title="简介"></a>简介</h1><p>相信大家都在高中接触过超几何分布. 超几何分布是<strong>不放回</strong>的, 该分布给出了抽取 $n$ 次后抽到次品次数的概率. 如果一个随机变量 $X$ 服从于超几何分布, 那么有<br>$$<br>P(X=k)=\frac{C_{N}^kC_{M-N}^{n-k}}{C_{M}^{n}}<br>$$<br>其中 $k$ 为抽取到的次品数, $M$ 为总数, $N$ 为总次品数.</p>
<h1 id="问题"><a href="#问题" class="headerlink" title="问题"></a>问题</h1><p>生活中很多变量都服从于超几何分布, 我们举个例子</p>
<blockquote>
<p>假设你们班要选出 $10$ 个人参加某个活动, 但是想要参加活动的人却有 $40$ 个人, 那么就要采取抽纸条的方式来选出参加活动的人. 准备 $40$ 张纸条, 其中有 $10$ 张是被标记了的. $40$ 个候选人每人抽取一张纸条, 抽到被标记纸条的人就可以参加活动.</p>
</blockquote>
<p>当已经有 $n$ 个人抽取了之后, $n$ 个人里面有 $X$ 个人可以参加活动, 这个 $X$ 就服从于超几何分布. 现在我们想要讨论的问题是: <strong>这样的抽取公平吗? 先抽和后抽对你抽到标记纸条的概率有没有影响?</strong> 再抽象一点, 即<strong>如果第 $n$ 次抽取时, 前面 $n-1$ 次抽取的结果都未知, 那么这次抽取抽到次品的概率是多少?</strong></p>
<p>直觉告诉我们, 应该是公平的, 也就是说在这样一个无返回抽样中, 无论先后, 抽取到次品的概率都是一样的, 并且这个值应该就是 $N/M$ .</p>
<h1 id="证明"><a href="#证明" class="headerlink" title="证明"></a>证明</h1><p>我们先说明几个数学符号, 以便后续推导</p>
<ul>
<li>$N$, $M$ 分别为次品数和总数</li>
<li>$X_1,X_2,\dots,X_n$ 分别为第 $n$ 次抽取抽取到的次品个数 ( $X_n$ 服从两点分布, 即其值要么 $0$ 要么 $1$)</li>
</ul>
<p>首先, 我们可以立即得出, $P(X_1=1) = \frac{N}{M}$ . 然后我们尝试计算 $P(X_2=1)$<br>$$<br>\begin{aligned}<br>P(X_2=1)&amp;=P(X_2=1,X_1=1)+P(X_2=1,X_1=0)\\<br>&amp;=P(X_2=1\mid X_1=1)P(X_1=1)+P(X_2=1\mid X_1=0)P(X_1=0)\\<br>&amp;=\frac{N}{M}\frac{N-1}{M-1}+\frac{M-N}{M}\frac{N}{M-1}\\<br>&amp;=\frac{N(M-N+N-1)}{M(M-1)}\\<br>&amp;=\frac{N}{M}<br>\end{aligned}<br>$$<br>果然, $P(X_2=1)$ 也等于 $\frac{N}{M}$ .</p>
<p>不过… 好像还是没有什么思路, 那我们继续计算 $P(X_3=1)$<br>$$<br>\begin{aligned}<br>P(X_3=1)&amp;=\sum_{X_1}\sum_{X_2}P(X_3,X_2,X_1)\\<br>&amp;=\sum_{X_1}\sum_{X_2}P(X_3\mid X_2,X_1)P(X_2\mid X_1)P(X_1)\\<br>&amp;=\left(\frac{N}{M}\frac{N-1}{M-1}\frac{N-2}{M-2}+\frac{N}{M}\frac{M-N}{M-1}\frac{N-1}{M-2}\right)\\&amp;+\left(\frac{M-N}{M}\frac{N}{M-1}\frac{N-1}{M-2}+\frac{M-N}{M}\frac{M-N-1}{M-1}\frac{N}{M-2}\right)<br>\end{aligned}<br>$$<br>并且有<br>$$<br>\frac{N}{M}\frac{N-1}{M-1}\frac{N-2}{M-2}+\frac{N}{M}\frac{M-N}{M-1}\frac{N-1}{M-2}=\frac{N}{M}\frac{N-1}{M-1}\\<br>\frac{M-N}{M}\frac{N}{M-1}\frac{N-1}{M-2}+\frac{M-N}{M}\frac{M-N-1}{M-1}\frac{N}{M-2}=\frac{M-N}{M}\frac{N}{M-1}<br>$$<br>于是同样的 $P(X_3=1)=\frac{N}{M}$ .</p>
<p>现在我们思考 $P(X_3=1)$ 与 $P(X_2=1)$ , $P(X_2=1)$ 与 $P(X_1=1)$ 之间有什么关系. 我们想象有很多轨迹, 每条轨迹最终都可以到达 $X_2=1$ 这种情况, 而将所有的能达到 $X_2=1$ 的轨迹产生的概率相加, 就是 $P(X_2=1)$ . $\frac{N}{M}\frac{N-1}{M-1}$ (第一次抽中, 第二次抽中) 和 $\frac{M-N}{M}\frac{N}{M-1}$ (第一次没抽中, 第二次抽中) 最终都可以达到<strong>第二次抽中</strong>这个结果 (也就是 $X_2=1$), 而也就只有这两条轨迹, 因此它们的概率相加便是 $P(X_2)=1$ .</p>
<p>类似的, 我们考虑 $X_3=1$ 的各个轨迹. $X_3=1$ 的轨迹和 $X_2=1$ 的轨迹有没有什么关联? 当然有, $X_3=1$ 的轨迹可以由 $X_2=1$ 的轨迹 “派生” 出来. 每一条 $X_2=1$ 轨迹都可以派生为两条 $X_3=1$ 轨迹, 比如说 $\frac{N}{M}\frac{N-1}{M-1}$ (第一次抽中, 第二次抽中), 这一条 $X_2=1$ 轨迹, 可以派生出 $\frac{N}{M}\frac{N-1}{M-1}\frac{N-2}{M-2}$ (第一次抽中, 第二次抽中, 第三次抽中) 与 $\frac{N}{M}\frac{(M-1)-(N-1)}{M-1}\frac{N-1}{M-2}$ (第一次抽中, 第二次没抽中, 第三次抽中) 两条轨迹, 即第二次抽中与没抽中的区别. 而 $X_4=1$ 的轨迹与 $X_3=1$ 的轨迹, $X_5=1$ 的轨迹与 $X_4=1$ 的轨迹, 乃至 $X_n=1$ 的轨迹与 $X_{n-1}=1$ 的轨迹之间, 都有这种派生关系. 如果能够证明, 一条轨迹出现的概率等于其派生出的两条轨迹出现的概率之和 (前面的计算已经看出在 $n=1,2$ 时满足这种关系), 那么就能够用数学归纳法证明每次抽取抽到次品的概率是相同的. 于是就要证明<br>$$<br>\cdots \frac{N-a}{M-b} = \cdots\frac{N-a}{M-b}\frac{N-a-1}{M-b-1}+\cdots\frac{(M-b)-(N-a)}{M-b}\frac{N-a}{M-b-1}<br>$$<br>由于 $X_n=1$ 的某条轨迹派生出 $X_{n+1}=1$ 的轨迹的过程中, 轨迹前面 $n-1$ 个动作 (即是否抽取到) 是不变的, 因此用 $\cdots$ 代替. 上面的式子显然成立, 于是我们就证明了 $P(X_n=1)=P(X_{n-1}=1)=\dots= P(X_1=1)=\frac{N}{M}$ . 也就是说, 游戏是公平的.</p>
<h1 id="应用"><a href="#应用" class="headerlink" title="应用"></a>应用</h1><p>据此, 我们可以轻松得出, 超几何分布的期望和二项分布是类似的. 即如果 $X$ 服从于超几何分布, 那么如果抽取 $n$ 次, $E(X)=n\frac{N}{M}=np$ (将 $\frac{N}{M}$ 看成一个概率 $p$ ).</p>
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